Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.io.*;
import java.math.BigInteger;
import java.util.*;

public class Program {

  static BufferedReader in;
  static PrintWriter out;
  static StringTokenizer tok = new StringTokenizer("");

  public static void main(String[] args) throws IOException {
    String filename = "";
    if (filename.isEmpty()) {
      in = new BufferedReader(new InputStreamReader(System.in));
      out = new PrintWriter(System.out);
    } else {
      in = new BufferedReader(new FileReader(filename + ".in"));
      out = new PrintWriter(filename + ".out");
    }
    solve();
    out.close();
  }

  static String readString() throws IOException {
    while (!tok.hasMoreTokens()) {
      tok = new StringTokenizer(in.readLine());
    }
    return tok.nextToken();
  }

  static int readInt() throws IOException {
    return Integer.parseInt(readString());
  }

  static long readLong() throws IOException {
    return Long.parseLong(readString());
  }

  private static void solve() throws IOException {
    int n = readInt();
    int m = readInt();
    int[][] matrix = new int[n][m];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        matrix[i][j] = readInt();
      }
    }
    List<String> ans = new ArrayList<>();
    if(n < m){
      for (int i = 0; i < n; i++) {
        int min = Integer.MAX_VALUE;
        for (int j = 0; j < m; j++) {
          min = Math.min(min, matrix[i][j]);
        }
        for (int j = 0; j < min; j++) {
          ans.add("row " + (i + 1));
        }
        for (int j = 0; j < m; j++) {
          matrix[i][j] -= min;
        }
      }
      for (int j = 0; j < m; j++) {
        int min = Integer.MAX_VALUE;
        for (int i = 0; i < n; i++) {
          min = Math.min(min, matrix[i][j]);
        }
        for (int i = 0; i < min; i++) {
          ans.add("col " + (j + 1));
        }
        for (int i = 0; i < n; i++) {
          matrix[i][j] -= min;
        }
      }
    } else {
      for (int j = 0; j < m; j++) {
        int min = Integer.MAX_VALUE;
        for (int i = 0; i < n; i++) {
          min = Math.min(min, matrix[i][j]);
        }
        for (int i = 0; i < min; i++) {
          ans.add("col " + (j + 1));
        }
        for (int i = 0; i < n; i++) {
          matrix[i][j] -= min;
        }
      }
      for (int i = 0; i < n; i++) {
        int min = Integer.MAX_VALUE;
        for (int j = 0; j < m; j++) {
          min = Math.min(min, matrix[i][j]);
        }
        for (int j = 0; j < min; j++) {
          ans.add("row " + (i + 1));
        }
        for (int j = 0; j < m; j++) {
          matrix[i][j] -= min;
        }
      }

    }

    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        if(matrix[i][j] > 0){
          out.print(-1);
          return;
        }
      }
    }
    out.println(ans.size());
    for(String s : ans){
      out.println(s);
    }
  }


}